Question

1. (28 Points) Symphony halls are designed to reduce the number of locations that produce intensity minima from reflections by the wall. In this problem you are asked to compute the frequencies that produce intensity minima if the room has bare concrete walls. Imagine a source of sound (S) that is played at the front stage of a symphony hall. A listener (L) is standing next to the cheap seats at the back of the hall. For simplicity let's assume that the speed of sound in air is 300 m/s and that the speed of sound in concrete is 1500 m/s. Both the source and the listener are located at the midpoint between the two concrete walls. The two paths that the sound will take are shown in the diagram (assume that the sound strikes the concrete at the 40 m mark) For wave 2 (the wave that strikes the concrete) is there a half wavelength phase shift? Explain. (Hint: Recall that thephase shift of light will occur if a. ni < n2. This can be rewritten in terms of the velocities of the wave.) b. What is the critical angle for total internal reflection and does sound wave 2 exceed the critical angle? What is the path length difference between the two waves? Write a formula for destructive interference in terms of the path length difference or. Determine the 3 lowest audible frequencies that produce complete destructive interference. Recall that audible frequencies are between 20 Hz and 20 kHz. c. d. e. 80 m Concrete Concrete

Answer 1

a) The velocity of the sound in air is
air 330(m/s) 12r
and  velocity of the sound in concrete is
Uconcrete 1500(m/s)

If the refractive index offered by the air for the sound wave is $n_1$ and refractive index offered by the concrete for the sound wave is $n_2$ then we find $\frac{n_2}{n_1}=\frac{v_{air}}{v_{concrete}}=\frac{300}{1500}=\frac{1}{5}\Rightarrow n_1>n_2$

According to stokes law when a wave (sound) moves from slow medium (air) or denser medium to faster medium (concrete ) or rarer medium and reflects back to slow medium (air), the wave will travel extra path of $\lambda/2$.

In our case, there will be half wavelength phase shift , when the sound wave reflects back to the air by hitting the the air concrete interface.

b) if the critical angle for total internal reflection is $\theta_c$ , then from Snells Law, we get

$n_{air}sin\theta_c= n_{concrete}sin 90^0$

Or

$sin\theta_c=\frac{n_2}{n_1}=\frac{1}{5}\Rightarrow \theta_c=sin^{-1}(\frac{1}{5})=11.54^0$

from geometry, we find

80 mo Concrete 30m 40m Concrete

$tan\theta=\frac{40}{30}\Rightarrow \theta=tan^{-1}(\frac{4}{3})=53.13^0$

Hence we can conclude sound wave 2 exceeds the critical angle.

c)

80 mo Concrete 30m 40m Concrete

The Path length difference between two waves is

$\delta_r= (SO+OL)+\frac{\lambda}{2}-SL$ [ $\lambda/2$ added because of Stokes Law]

From geometry we find

$\delta_r= [\frac{\lambda}{2}+\sqrt{40^2+30^2}+\sqrt{40^2+30^2}-80]m$

Z0im

d) For destructive interference the path length should be half odd multiple of the wavelength .

.ie  $\delta_r =\frac{n \lambda}{2}$ [ where n= 1, 3, 5, 7,......]

Answer 2

m 20 m d. The formula for destructive interference is given as follows "",""error"":null,""modified"":1558432543,""source"":""automated""}"